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3 years ago

6

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is normally distributed with A random s

ample of 12 sample specimens has a mean compressive strength of psi. Round your answers to 1 decimal place. (a) Calculate the 95% two-sided confidence interval on the true mean compressive strength of concrete. Enter your answer; 95% confidence interval, lower bound Enter your answer; 95% confidence interval, upper bound (b) Calculate the 99% two-sided confidence interval on the true mean compressive strength of concrete.

1 answer:

wel3 years ago

3 0

Answer:

95%: (3278.354 ; 3270.083)

99% : (3221.646 ; 3278.354)

Step-by-step explanation:

Given :

Sample size, n = 12

Mean, xbar = 3250

Sample standard deviation = √1000

The 95% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.05, df=12-1 = 11 ;

Tcritical at 95% = 2.20

Hence,

Margin of Error = (2.20 * √1000/√12) = 20.083

Confidence interval : 3250 ± 20.083

Lower boundary = 3250 - 20.083 = 3229.917

Upper boundary = 3250 + 20.083 = 3270.083

2.)

The 99% confidence interval :

Mean ± Margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 0.01, df=12-1 = 11 ;

Tcritical at 99% = 3.106

Hence,

Margin of Error = (3.106 * √1000/√12) = 28.354

Confidence interval : 3250 ± 28.354

Lower boundary = 3250 - 28.354 = 3221.646

Upper boundary = 3250 + 28.354 = 3278.354

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Military radar and missile detection systems are designed to warn a coutry of an enemy attack. Assume that a particular detectio

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Answer:

0.96 = 96% probability that at least one of them detect an enemy attack.

Step-by-step explanation:

For each radar, there are only two possible outcomes. Either it detects the attack, or it does not. The missiles are operated independently, which means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Assume that a particular detection system has a 0.80 probability of detecting a missile attack.

This means that $p = 0.8$

If two military radars are installed in two different areas and they operate independently, the probability that at least one of them detect an enemy attack is

This is $P(X \geq 1)$ when $n = 2$ . So

$P(X \geq 1) = 1 - P(X = 0)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{2,0}.(0.8)^{0}.(0.2)^{2} = 0.04$

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.04 = 0.96$

0.96 = 96% probability that at least one of them detect an enemy attack.

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3 years ago

The area of this circle is 42π m².

kondor19780726 [428]

A full circle is 360°, so a 60° circle is 1/6th of the area.

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A car used 11.4 gallons of gas on a 324.9 mile trip. How many miles can this car travel on one gallon of gas?

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324.9 : 11,4 = 28.5
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3 years ago

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A manufacturer of small appliances purchases plastic handles for coffeepots from an outside vendor. If a handle is cracked, it i

Rus_ich [418]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$

And rounded up we got:

$n\approx 97$

Step-by-step explanation:

Data given and previous concepts

ME=0.1 represent the margin of error desired

Confidence =0.95 or 95%

$\alpha=0.05$ represent the significance level

z represent the quantile from the normal standard distribution

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

Since we don't have a prior estimate for the proportion $\hat p$ we can use as estimator 0.5 and replacing into equation (b) the values from part a we got:

$n=\frac{0.5(1-0.5)}{(\frac{0.1}{1.96})^2}=96.04$

And rounded up we have that n=97

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4 years ago

The window shown is the shape of a semicircle with a radius of 6 feet. The distance from F to E is 3 feet and the measure of = 4

Sergeu [11.5K]

Answer:

The area of the glass in region BCIH is 11 to the nearest feet²

Step-by-step explanation:

* Lets explain the figure

- The window is a semicircle with center G and radius 6 feet

- There is a small semicircle with center G and radius GF

∵ GE is 6 feet and EF is 3 feet

∵ GE = GF + FE

∴ 6 = GF + 3 ⇒ subtract 3 from both sides

∴ 3 = GF

∴ The radius of the small semicircle is 3 feet

∵ m∠BGC = 45°

- The area of sector BGC is part of the area of the semicircle

∵ The area of semi-circle is 1/2 π r²

∵ The measure of the central angle of the semicircle is 180°

∵ The measure of the central angle of the sector BGC is 45°

∴ The sector = 45°/180° = 1/4 of the semi-circle

∴ The area of the sector is 1/4 the area of the semicircle

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∵ r = 6 feet

∴ The area of the semicircle = 1/2 π (6)² = 1/2 π (36) = 18 π feet²

∴ Area of the sector = 1/4 (18 π) = 4.5 π feet²

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∴ Area of the sector HGI= 1/4 (4.5 π) = 1.125 π feet²

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area of sector BGC and the area of the sector HGI

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3 years ago

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